Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
-
$38.52\ l$
-
$43.74\ l$
-
$42.68\ l$
-
$39.85\ l$
-
$None\ of\ these$
B
Correct answer
Explanation
Initial milk = 60. After 3 replacements of 6L: Final milk = 60 * (1 - 6/60)^3 = 60 * (0.9)^3 = 60 * 0.729 = 43.74 L.
-
$54.23$litre
-
$54.26$litre
-
$52.26$litre
-
$52.48$litre
-
$14 \dfrac{2}{7} L$
-
$86 \dfrac{3}{7} L$
-
$187\dfrac{1}{7} L$
-
$55\dfrac{5}{7} L$
-
13 : 55
-
20 : 74
-
16 : 65
-
10 : 48
C
Correct answer
Explanation
The formula for the final concentration of the original liquid after n replacements is: Final = Initial * (1 - x/y)^n. Here, 1/3 is replaced, so 2/3 remains. After 4 operations, the ratio of dettol to total volume is (2/3)^4 = 16/81. Thus, dettol is 16 parts and water is 81 - 16 = 65 parts. The ratio is 16:65.
-
30 : 75
-
31 : 74
-
31 : 75
-
30 : 74
A
Correct answer
Explanation
Initial mixture: 100 L, ratio 3:1. Milk = 75 L, Water = 25 L. Add 200 L water: New water = 25 + 200 = 225 L. New ratio = 75 : 225 = 1 : 3.
-
$\displaystyle \frac{719}{371}$
-
$\displaystyle \frac{709}{291}$
-
$\displaystyle \frac{727}{273}$
-
$\displaystyle \frac{729}{271}$
D
Correct answer
Explanation
Initial paint = 20. After 3 replacements of 2L: Final paint = 20 * (1 - 2/20)^3 = 20 * (0.9)^3 = 20 * 0.729 = 14.58. Oil = 20 - 14.58 = 5.42. Ratio Paint/Oil = 14.58 / 5.42 = 1458 / 542 = 729 / 271.
-
25
-
30
-
45
-
Cannot be determined
B
Correct answer
Explanation
Let initial petrol be 3x and kerosene be 2x. Total = 5x. Removing 10L removes 6L petrol and 4L kerosene. Remaining: (3x-6) petrol and (2x-4+10) kerosene. Ratio (3x-6)/(2x+6) = 2/3. 9x - 18 = 4x + 12, so 5x = 30.
-
$\dfrac 19$
-
$\dfrac 39$
-
$\dfrac 59$
-
$\dfrac 79$
-
$21\ kg$
-
$16\ kg$
-
$4\ kg$
-
$7\ kg$
A
Correct answer
Explanation
Use allegation: (Price_tea - Price_mix) / (Price_mix - Price_chicory) = Quantity_chicory / Quantity_tea. (10 - 6.5) / (6.5 - 4) = x / 15. 3.5 / 2.5 = x / 15. 1.4 = x / 15. x = 21.
-
$20\ Litres$
-
$30\ Litres$
-
$40\ Litres$
-
$60\ Litres$
D
Correct answer
Explanation
Initial milk is 40L and water is 20L. Let x be the water added: 40 / (20 + x) = 1 / 2. Solving for x gives 80 = 20 + x, so x = 60.
B
Correct answer
Explanation
Initial mixture: 56 liters, ratio 5:2. Milk = 40, Water = 16. Let x be water added. New ratio: 40 / (16+x) = 4/5. 200 = 64 + 4x. 136 = 4x. x = 34.
-
72.9 L
-
65.61 L
-
34.39 L3
-
81 L
B
Correct answer
Explanation
The formula for remaining amount is A = P * (1 - r/P)^n. Here P=100, r=10, n=4. A = 100 * (0.9)^4 = 100 * 0.6561 = 65.61 L.
-
$43:96$
-
$438:962$
-
$348:962$
-
$481:219$
D
Correct answer
Explanation
Vessel volumes are 2x, 3x, 5x. Milk/Water ratios: V1 (1:3) -> 0.5 milk, 1.5 water; V2 (2:3) -> 1.2 milk, 1.8 water; V3 (2:5) -> 1.428 milk, 3.571 water. Summing these and simplifying gives the ratio 481:219.